Difference between revisions of "2007 AMC 8 Problems/Problem 1"

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So, the answer is <math>\boxed{D}</math>
 
So, the answer is <math>\boxed{D}</math>
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==See Also==
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{{AMC8 box|year=2007|before=First<br />Question|num-a=2}}
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[[Category:Introductory Algebra Problems]]

Revision as of 14:03, 18 May 2011

Problem

Theresa's parents have agreed to buy her tickets to see her favorite band if she spends an average of $10$ hours per week helping around the house for $6$ weeks. For the first $5$ weeks she helps around the house for $8$, $11$, $7$, $12$ and $10$ hours. How many hours must she work for the final week to earn the tickets?

$\mathrm{(A)}\ 9 \qquad\mathrm{(B)}\ 10 \qquad\mathrm{(C)}\ 11 \qquad\mathrm{(D)}\ 12 \qquad\mathrm{(E)}\ 13$

Solution

Let $x$ be the number of hours she must work for the final week. We are looking for the average, so $\frac{8 + 11 + 7 + 12 + 10 + x}{6} = 10$

Solving gives:

$\frac{48 + x}{6} = 10$

$48 + x = 60$

$x = 12$

So, the answer is $\boxed{D}$

See Also

2007 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
First
Question
Followed by
Problem 2
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All AJHSME/AMC 8 Problems and Solutions